Examine these two queries and their

单选题
Examine these two queries and their output:
SELECT deptno, dname FROM dept;
DeptnoDname
10account ing
20research
30sales
40operations
SELECT ename, job, deptno FROM emp ORDER BY deptno;
Enamejobdeptno
CLARKMANAGER10
KINGPRESIDENT10
MILLERCLERK10
JONESMANAGER20
Now examine this query:
SELECT ename, dname
FROM emp CROSS JOIN dept WHERE job = 'MANAGER'
AND dept.deptno IN (10, 20) ;

A、 64

B、 6

C、 3

D、 12

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单选题
In your session, the NLS._DAE_FORMAT is DD- MM- YYYY.There are 86400 seconds in a day.Examine
This result:
DATE
02-JAN-2020
Which statement returns this?

A、SELECT TO_ CHAR(TO_ DATE(‘29-10-2019’) +INTERVAL ‘2’; MONTH + INTERVAL ‘5’; DAY -
INTERVAL ‘86410’ SECOND, ‘ DD-MON-YYYY’) AS "date"
FROM DUAL;

B、SELECT TO_ CHAR(TO_ DATE(‘29-10-2019’) + INTERVAL ‘3’ MONTH + INTERVAL ‘7’ DAY -
INTERVAL ‘360’ SECOND, ‘ DD-MON-YYYY’) AS "date"
FROM DUAL;

C、SELECT To CHAR(TO _DATE(‘29-10-2019’) + INTERVAL ‘2’ NONTH + INTERVAL ‘5’ DAY
INEERVAL ‘120’ SECOND, ‘ DD-MON-YYY) AS "date"
FROM DUAL;

D、SELECT-TO_CHAR(TO _DATE(‘29-10-2019’+ INTERVAL ‘2’ MONTH+INTERVAL ‘6’ DAY-
INTERVAL ‘120’ SECOND, ‘DD-MON-YY’) AS "daTe"
FROM DUAL;

E、SELECT-TO_CHAR(TO _DATE(‘29-10-2019’+ INTERVAL ‘2’ MONTH+INTERVAL ‘4’ DAY-
INTERVAL ‘120’ SECOND, ‘DD-MON-YY’) AS "daTe"
FROM DUAL;

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单选题
Choose the best answer.
Examine the description of the EMPLOYEES table:
NameNullType
----------------------------------------------------------
EMP_IDNOT NULNUMBER
EMP_NAMEVARCHAR2 (40)
DEPT_IDNUMBER(2)
SALARYNUMBER(8,2)
JOIN_DATEDATE
Which query is valid?

A、SELECT dept_id, join_date,SUM(salary) FROM employees GROUP BY dept_id, join_date;

B、SELECT depe_id,join_date,SUM(salary) FROM employees GROUP BY dept_id:

C、SELECT dept_id,MAX(AVG(salary)) FROM employees GROUP BY dept_id;

D、SELECT dept_id,AVG(MAX(salary)) FROM employees GROUP BY dapt_id;

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单选题
Examine the description of the EMPLOYEES table:
NameNull?Type
--------------------------------------------
EMPLOYEE IDNOT NULLNUMBER (4)
LAST NAMENOT NULLVARCHAR2 (100)
SALARYNOTNOT NULLNUMBER (6,2)
DEPARTMENT_ IDNOT NULLNUMBER(4)
Examine this query:
1 SETECTe。 last_ name,
2e。 salary,
3a。 avg_ sal
4FROM employees e
5WHERE e。 salary > (SELECT AVG (a。 salary) AS avg _ _sal
6FROMemployees a
7WHERE a。 department_ 1d = e。epartment。 1d)
8ORDER BY e。 last_ name;
Which line produces an error?

A、Line 7

B、Line 8

C、Line 3

D、Line 5

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单选题
NameNull?Type
-------------------------------------------------------------------------------
EMP_NONOT NULLNUMBER(4)
LAST_NAMEVARCHAR2(100)
HIRE_DATEDATE
SALARYNUMBER(8,2)
For each employee in department 90 you want to display
1 their last name
2 the number of complete weeks they have been employed
The output must be sorted by the number of weeks, starting with the longest serving employee
First.Which statement will accomplish this?

A、SELECT last_name, TRUNC( (SYSDATE - hire_ date) 1 7) AS tenure
FROM employees
WHERE department_ id = 90
ORDER BY tenure ;

B、SELECT last_name, ROUND( (SYSDATE - hire_ date) 1 7) AS tenure
FROM employees
WHERE department_ id = 90
ORDER BY tenure ;

C、SELECT last_name, ROUND( (SYSDATE - hire_ date) 17) AS tenure
FROM employees
WHERE department_ id = 90
ORDER BY tenure DESC;

D、SELECT last_name, TRUNC ( (SYSDATE - - hire_ date) 1 7) AS tenure
FROM employees
WHERE department_id = 90
ORDER BY tenure DESC;

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单选题
Examine the data in the EMPLOYEES table:
EMPLOYEE_ID LAUT NANE MONTHLY SALARY MONTTHLY COMEHISSIOM PCT
101Kochhar24000<NULL>
102Ernst17000.5
103Rajs21000.2
104LORENTZ25000<NULL>
105Morris12000<NULL>
Which statement will compute the total annual compensation for each employee?

A、SELECT last _ NAME (monthly_ salary + monthly _commission _ pct) * 12 AS annual_ comp
FROM employees;

B、select last _ name, (monthly_ salary * 12) + (monthly_ salary * 12 *monthly_ commission_ pct)
AS annual_ camp FROM employees

C、SELECT last _ name, (monthly_ salary * 12) + (monthly_ salary * 12 * NVL (monthly_ commission _
Pct, 0)) AS annual _comp

D、SELECT last _ name, (monthly _ salary * 12) + (monthly_ commission _ pct * 12) AS
FROM employees:

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单选题
You must find the number of employees whose salary is lower than employee 110.
Which statement fails to do this?

A、SELECT COUNT (*)
FROM employees
JOIN employees a
ON e。 salary< a。 salary
WHERE a。employee_ id= 110;

B、SELECT COUNT (* )
FROM employees
WHERE salary < (SELECT salary FROM employees WHERE employee 业id =
110) ;

C、SELECT COUNT (*)
FROM employees e
JOIN (SELECT salary FROM employees WHERE employee_ id= 110) a
ON e。salary< a。salary;

D、SELECT COUNT (* )
FROM employees e
WHERE e。salary < (SELECT a。salary FROM employees a WHERE e。employee_ id = 110);

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单选题
Examine the data in the EMPLOYEES table:
EMPLOYEE ID LAST NAME MONTHLY SATARY MONTHLY COMMISSION PCT
101Kochhar24000<NULL>
102Ernst17000.5
103Rajs21000.2
104Lorentz25000<NULL>
105Morris12000<NULL >
Which statement will compute the total annual compensation for each employee?

A、SELECT last name,
(monthly salary*12) + (monthly_commission_pct * 12) AS
Annual comp
FROM employees
;

B、SELECT last_ name (monthly_ salary+ monthly_ commission _ pct) *12 AS annual_
FROM employees ;

C、SELECT last name, (monthly_ salary *12) + (monthly_ salary * 12 * NVL
(monthly commission pct,0) ) As annual _ comp
FROM employees;

D、SELECT last_ name, monthly_ salary*12) + (monthly_ salary * 12 * Monthly commission _Pct) AS
Annual_ comp
FROM employees;

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单选题
Choose the best answer
The CUSTOMERS table has a CUST_LAST_NAME column of data type VARCHAR2.
The table has two rows whose COST_LAST_MANE values are Anderson and Ausson.
Which query produces output for CUST_LAST_SAME containing Oder for the first row and Aus for the
Second?

A、SELECT REPLACE (REPLACE(cust_last_name,'son',''),'An','O') FROM customers;

B、SELECT REPLACE (TRIM(TRALING 'son' FROM cust_last_name), 'An','O') FROM customers;

C、SELECT INITCAP (REPLACE(TRIM('son' FROM cust_last_name),'An','O')) FROM customers;

D、SELECT REPLACE (SUBSTR(cust_last_name,-3), 'An', 'O') FROM customers;

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